package subarraysDivByK;

/**
 * @author ffzs
 * @describe  974. 和可被 K 整除的子数组
 *
 * 给定一个整数数组 A，返回其中元素之和可被 K 整除的（连续、非空）子数组的数目。
 *
 * 示例：
 *
 * 输入：A = [4,5,0,-2,-3,1], K = 5
 * 输出：7
 * 解释：
 * 有 7 个子数组满足其元素之和可被 K = 5 整除：
 * [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
 *
 * 提示：
 *
 * 1 <= A.length <= 30000
 * -10000 <= A[i] <= 10000
 * 2 <= K <= 10000
 *
 * @date 2020/5/27
 */
public class Solution {
    public static int subarraysDivByK(int[] A, int K) {
        int sum = 0, count = 0;
        int[] tmp = new int[K];
        tmp[0] = 1;
        for (int value : A) {
            sum += value;
            int remainder = sum < 0 ? sum%K + K: sum%K;
            count += tmp[remainder];
            tmp[remainder]++;
        }

        return count;
    }

    public static void main(String[] args) {
        int[] A = {4, 5, 0, -2, -3, 1};
        int K = 5;
        System.out.println(subarraysDivByK(A, K));
    }
}
